Pushforward of semi-stable sheaves under finite field extension Let $k$ be a field of characteristic zero and $X$ be a non-singular rationally connected variety over $k$. Let $L$ be a finite field extension of $k$. This induces a proper morphism $p:X_L \to X_k$. Is it true that for any torsion-free semi-stable sheaf $E$ on $X_L$, the coherent sheaf $p_*E$ is semi-stable on $X_k$?
 A: Assuming what you mean is that $X$ is projective and $\mathcal E$ is Gieseker-semistable, the answer is yes.


Lemma. Let $X$ be a projective $k$-scheme, and let $\mathcal E$ be a semistable sheaf on $X$. Then $\mathcal E_{\bar k}$ is semistable on $X_{\bar k}$.


Proof. Let $\mathcal F \subseteq \mathcal E_{\bar k}$ be a proper nonzero subsheaf, and assume that $\mu(\mathcal F) > \mu(\mathcal E)$. Then $\mathcal F$ is defined over some finite extension $k \subseteq \ell$. Let $p \colon X_\ell \to X$ be the projection. Then $p_* \mathcal F \subseteq p_* \mathcal E_\ell$ is a subsheaf, with $\mu(p_*\mathcal F) > \mu(p_* \mathcal E_\ell)$. But $p_* \mathcal E_\ell = p_* p^* \mathcal E = \mathcal E^{[\ell:k]}$ is semistable, so such subsheaves cannot exist. $\square$
Remark. The same statement with semistable replaced by stable is false. See for example this post.
Remark. Conversely, it follows from uniqueness of the Harder–Narasimhan filtration that if $\mathcal E_{\bar k}$ is semistable, then so is $\mathcal E$. Indeed, uniqueness of the HN filtration implies (using the lemma above) that it commutes with base change of the field.


Corollary. Let $X$ be a projective $k$-scheme, let $k\subseteq \ell$ be a finite separable extension, and let $\mathcal E$ be a semistable sheaf on $X_\ell$. Then $p_* \mathcal E$ is semistable.


Proof. By the remark, it suffices to show that $(p_* \mathcal E)_{\bar k}$ is semistable. Because $\ell \otimes_k \bar k \cong \bar k^{[\ell:k]}$, the base change of $p \colon X_\ell \to X$ to $\bar k$ looks like
$$p_{\bar k} \colon \coprod_{\sigma \colon \ell \to \bar k} X_{\bar k} \to X_{\bar k}.$$
By the lemma above, the base change $\mathcal E_{\bar \ell}$ is semistable. Under $p_{\bar k}$, we are merely taking the sum of the Galois conjugates of $\mathcal E_{\bar \ell}$. It is straightforward to check that each of these Galois conjugates is semistable of the same slope, hence their sum is semistable. $\square$
